3.1 If a s 2(mod4) then the derived group of

F = < R,S I R^ = ~ (RS)^RS^R^S^ *= 1 > is a perfect group and has index 4 in F,

Proof: Let H = < R, SRS'*, S^RS'^, S^RS’^ > be a subgroup of F. We can define cosets H = 1, IS = 2, 2S = 3, 3S = 4. No collapses occur. We now consider coset representatives rather than cosets. Let us put A = R , B = SRS"\ C = S%S"^, D = S^RS’^. The coset table and coset representative table are respectively ;

R S R-^ S ■1 R S R"^

1 1 2 1 4 1 A1 2 A-^1 4

2 2 3 2 1 2 B2 3 B-l2 1

3 3 4 3 2 3 C3 4 C-I3 2

4 4 1 4 3 4 D4 1 D-M 3

From the first relation of F we get the first four nontrivial relations for H and the 3 relation of F we get the last four nontrivial relations for H as follows :

(1) A4 = 1, (5) (ABCD)M/4ABCB2 = 1,

(2) b4 = _{1, (6)} (BCDA)(^-2)/4bCDC2= 1,

(3) C4 = 1, (7) (DABC)(^-7)/4dABA2 = 1,

Abelianizing the relations of F we get I F/F' I = 4 so the index of F' is 4 but on the other hand the generators of H belong to F' (to see take F/F' = < R,S I R =1, = 1 >

so R eF' ) therefore H = F'. Now we are going to show that F’ = H is a perfect group. In order to do this we have to consider four cases for (a-2)/4.

Case (a-2)/4 s 0(mod4): Let us assume that (a-2)/4 = 0(mod4). Abelianizing the relations of H and using first four relations of H/H’ we get the following relations for H/H' :

(I) A'* = 1. (HI) C* = 1, (V) ACB3 = 1, (VII) DBA3 = 1, (n> S'* = I, (IV) O'* = 1, (VI) b d c3 = I, (v n i) c a d^ = i

From relation (V) and (VIII) we get B^ = ...( i) . From relation (VI) and (VII) we get A^ = .... (ii). Considering first four relations of H/H' and using (i), (ii) we get

A = C, B = D. Eliminating C and D we get

H/H’ = < A, B I A^ = B"^ = A^B^ = B^A^ = 1 >, which can easily be seen to be the trivial group. Therefore H = F’ is perfect.

Case (a-2)/4 s l(mod4); Let us assume that (a-2)/4 s l(mod4). Abelianizing the relations of H and using the first four relations of H/H’ we get the following relations for H/H’ :

(I ) A 4 = 1 , ( n i ) C 4 = i _ (V) A^C^D = I, (v n ) B^CD^ = 1,

(II) S'* = 1, (IV) O'* = I, (VI) A S W = 1, (Vni) A^BC^ = 1

From relation (V) and (VIII) we get B = D. From relations (VI) and (VII) we get A = C. Eliminating C and D we get the trivial group. Therefore H = F’ is a perfect group.

Case (a-2)/4 s 2(mod4): Let us assume that (a-2)/4 = 2(mod4). Abelianizing the relations of H and using the first four relations of H/H' we get the following relations for H/H'.

(I) A'* = I, (HI) C = 1, (V) A%C3d2 = 1. (VII) AB^C^D^ = 1, ( n) B ' * =l, (IV)D'* = 1, (VI) A^B^CD^ = 1, (Vni) A^B^C^D = 1

From relation (V) and (VIII) we get B - D. From relation (VI) and (VII) we get A = C.

Eliminating C and D gives:

H/H' = < A, B I a4 = S'* = A^b3 = B^A^ = 1 >

which can easily be seen to be the trivial group. Therefore H = F' is perfect. Case (a-2)/4 s 3(mod4): Let us assume that (a-2)/4 s 3(mod4). Abelianizing the relations of H and using the first four relations of H/H' we get the following relations for H/H' :

(I) A‘* = l, (HI) C * = l, (V) b2d3 = 1, (VH) A^C^ = 1,

(II) B'* = 1, (IV) D'* = I. (VI) a 3 c 2 = I , (VIII) b3d 2= 1

From relation (V) and (VIII) we get B - D. From relation (VI) and (VII) we get A = C. Eliminating C and D we get the trivial group. Therefore H == F' is a perfect group.

Theorem 2.3.2 If a s l(mod4) then the derived group of

F = < R,S I R4 z= g4 = (RS)^RS^R^S^ = 1 > is a perfect group and has index 4 in F.

Proof: Let H = < S, RSR'*, R^SR'^, R^SR'^ > be a subgroup of F. We can define cosets H = 1, IR = 2, 2R = 3, 3R = 4. No collapses occur. We now consider coset representatives rather than cosets. Let us put A = S , B = RSR"\ C = R^SR'^, D = R^SR’^. The coset table and coset representative table are respectively :

R S R-^ R_______S_______R-^ S'^

1 2 1 4 1 1 2 A1 4 A "h

2 3 2 1 2 2 3 B2 1 B"^2

3 4 3 2 3 3 4 C3 2 C'^3

From the second relation of F we get the first four nontrivial relations for H and from the 3 relation of F we get the last four nontrivial relations for H as follows:

(1) A '*=l, (5) (BCDA)(a-l)/4BC3A3 = 1, (2) B'* = l, (6) (CDAB)(»-*)/4cd3b3 = 1, (3) C* = 1, (7) (DABC)(“-iy^DA3c3 = 1, (4) D '*= l, (8) (ABCD)(^-1)/‘*AB3 3 = 1.

Abelianizing the relations of F we get I F/F* I = 4 so the index of F* is 4. On the

other hand the generators of H belong to F’ so H = F*. Now we are going to show that

F* - H is a perfect group. In order to do this we have to consider four cases for (a-l)/4.

Case (a-l)/4 = 0(mocl4): Let us assume that (a-l)/4 = 0(mod4). Abelianizing the relations of H and using the first four relations of H/H* we get the following relations for H/H* :

(I) A'* = 1, (HI) = 1, (V) BC3a3 = 1, (VII) D a3c3 = I,

(H) b4 = 1, (IV) O'* = 1, (VI) c d3b3 = 1, (vn i) a b^d^ = i

From relation (V) and (VII) we get B = D. From relation (VI) and (VIII) we get A = C. Eliminating C and D we get the trivial group. Therefore H = F* is a perfect group.

Case (a-l)/4 s l(mod4): Let us assume that (a-l)/4 s l(mod4). Abelianizing the relations of H and using the first four relations of H/H* we get the following relations for H /H ':

(I) a4 = i, (HI) c* = i, (V) b2d = i,

(vn)

From relation (V) and (VII) we get B = D. From relation (VI) and (VIII) we get A = C. Eliminating C and D we get the trivial group. Therefore H = F' is a perfect group.

relations of H and using the first four relations of H/H* we get the following relations for H/H*:

(I) a4 = 1, (HI) e* = 1, (V) a b3c d2 = 1, (v n ) a b^c d^ = i,

(H) B'* = 1, (IV) O'* = 1, (VI) A ^B tfD = I, (VIII) A^BC^D = 1.

From relation (V) and (VII) we can get B = D. From relation (VI) and (VIII) we can get A = C. Eliminating C and D we get the trivial group. Therefore H = F* is a perfect group.

Case (a-l)/4 = 3(mod4); Let as assume that (a-l)/4 = 3(mod4). Abelianizing the relations of H and using the first four relations of H/H* we get the following relations for H/H* :

(I) A' *=l, (HI) C * = I , (V) A2c2d3 = 1, (VII) a2b3c2 = 1 ,

(Ü) B'* = l, (IV) 0 4 = 1 , (VI) A3b2d2 = 1 , (v n i) B^C^D^ = 1.

From relation (V) and (VII) we get B = D. From relation (VI) and (VIII) we get 1 A = C. Eliminating C and D we get the trivial group. Therefore H = F* is a perfect

group.

Theorem 2.3.3 If a = 0(mod4) then the derived group of j

F = < R,S 1 r4 = S4 = (RS)^RS^R^S^ = 1 > is a perfect group and has index 4 in F.

Proof: Let H = < S^R, RS^, S^RS"\ RSR'^S'^ > be a subgroup of F. We can define cosets H = 1, IS = 2, 2S = 3, 38 = 4. No collapses occur. We now consider coset representatives rather than cosets. Let us put A = S^R, B = RS^, C = S^RS'^ D = RSR'^S"^ The coset table and coset representative table are respectively :

R S R'^ S'^ _______ R S_______R'^ S'^

1 3 2 3 4 1 B3 2 A'^S 4

2 4 3 4 1 2 D-1B4 3 C M 1

3 1 4 1 2 3 A1 4 B 'h 2

From the first relation of F we get the first two nontrivial relations for H, the other two relations are redundant, and from the 3 relation of F we get the last four nontrivial relations for H as follows :

(1) (BA)2=1, (4) (D-]B^CA)='/'*D-lBAB = 1,

(2) (D-*BC)2=1, (5) (AD-1b2C)^''ACD-1b = 1,

(3) (BCAD-lB)a/4BD-*BC = l, (6) (CAD *B2)a/4CBA=l.

Abelianizing the relations of F we get I F/F' I = 4 so the index of F' is 4. On the other hand the generators of H belong to F' so H = F'. Now we are going to show that

F' = H is a perfect group. In order to do this we have to consider four cases for a/4. Case a/4 = 0(mod4): Let us assume that a/4 = 0(mod4). Abelianizing the relations of H and consider a/4 - 4k, k e Z . We get the following relations for H/H' :

(I) A2r2 = 1 (III) A4kB8k+2Q4k+lp)-4k-l = i (y) A4k+lB8k+lc4k+l£)-4k-l = % (II) = 1 (IV) A4k+lB8k+2c4kj)-4k-l = ^ (y i) A4k+lB8k+lc4k+l0-4k = j Using relations (I) and (II) we can simplify the relations of H/H' as

(i) A W = 1 (iii) B^CD-l = 1 (v) BCD-*A = 1

(Ü) bW d-2 = 1 (iv) b2d->A = 1 (vi) ABC=1.

From relation (v) and (vi) we get D - 1. From relation (3) and (4) we get A = C. Eliminating C we get H/H' = < A, B I A^B^ = AB^ = A^B = 1 > , which can easily be seen to be the trivial group. Therefore H = F* is a perfect group.

Case a/4 = l(mod4); Let us assume that a/4 = l(mod4). Abelianize the relations of H and consider a/4 = 4k+l, k eZ . We get the following relations for H /H '.

(I) A^B^ = 1 (IE) A4k+lB8k+4c4k+2j)-4k-2 = j (y) A4k+2g8k+3c4k+2p)-4k-2 = % (H) B^C^D-^ = 1 (iy)A4k+2B8k+4c4k+lp-4k-2 = i (y j ) A4k+2g8k+3c4k+2£)-4k-l = j Using relations (I) and (II) we simplify the relations of H/H' as

(i) A% 2 _ 1 (iii) b^A = 1 (v) B-* = 1 (ii) B W d-2 = 1 (iv) C* = 1 (vi) BC2d-1 = 1

From these relations it is easily seen that it is the trivial group. Therefore H = F' is

perfect.

Case a/4 = 2(mod4): Let us assume that a/4 = 2(mod4). Abelianize the relations of H and consider a/4 = 4k+2, k e Z . We get the following relations for H/H* :

(I) A^B^ = 1 (III) A4k+2g8k+6(;4k+3p-4k-3 = % (V) A4k+3g8k+5Q4k+3p)-4k-3 = j (E) B^C^D’^ = 1 (IV) A4k+3B8k+6c4k+2£)-4k-3 = i (VI ) A4k+3B8k+5c4k+3o-4k-2 Using relations (I) and (II) we simplify the relations of H/H* as

(i) = 1 (iii) e * D = 1 (v) ABCD-1 = 1

(ii) B W d-2 = 1 (iv) AB2d-1 = 1 (vi) ABC=1 .

It can easily be seen that it is the trivial group, therefore H = F* is perfect.

Case a/4 = 3(mod4): Let us assume that a/4 s 3(mod4). Abelianize the relations of H and consider a/4 = 4k+3, k e Z . We get the following relations for H/H* ;

(I) A^B^ = 1 (IE) A4k+3g8k+8(4k+42;).4k-4 = i (y) A4k+4B8k+7c4k+4p)-4k-4 = ^ (E) B^C^D-^ = 1 (IV)A4k+4B8k+8c4k+3p)-4k-4 = i (y j ) A4k+4g8k+7c4k+4D-4k.3 = j Using relations (I) and (II) we simplify the relations of H/H* as

(i) a2b2 = 1 (iii) b2a = 1 (v) B-1 = 1

(ii) b2c2d-2 = 1 (iv) C"l = l (vi) D B -1-1.

From these relations it can easily be seen that it is the trivial group. Therefore H = F* is perfect.